Wolfram Alpha is an excellent resource for napkin math – it handles most of the imperial to metric unit conversions, and frequently can return relevant facts from a natural language query. In this case it appears Wolfram Alpha’s facts about diesel fuel are slightly off; per this excellent fuel data sheet, diesel petroleum at STP (standard temperature and pressure) combusts for 35.94 MJ / liter.

Dividing the mechanical power – 8.3 hp or 6.2 kW – by the thermal power yields the engine’s thermal efficiency of 71%. This is about 40% higher than very efficient diesel engines at around 50%. Quite an astounding efficiency claim.

Carl Sagan frequently stated that “extraordinary claims require extraordinary evidence”; I have a lot of assumptions in flight here and I’d love to see a direct claim from Volkswagen that confirms or repudiates my napkin math. An alternate explanation is that the XL1’s fuel consumption rate is being misrepresented – for example by listing only the fuel consumed in a combined cycle test that uses both fuel and electrical energy.

If Volkswagen has indeed come up with a huge improvements in diesel efficiency, it would have impacts well beyond a couple hundred eco-supercars. Broad freight applications – 18 wheeler, train, ship – could potentially make a huge dent in worldwide diesel fuel consumption. But let’s see a more direct claim first.

How about an electric XL1?

Pull the diesel powertrain, slap the Nissan Leaf’s 24 kWh battery and motor in here and you’d have a MUCH faster vehicle (~6s 0-60). Weight would probably increase by a few hundred pounds, but power to travel at 60 mph should only change slightly. Assuming the same mechanical energy requirements and 90% battery to motor efficiency would give about 200 miles of range at 60 mph .. and an 80% charge in 30 minutes from CHAdeMO. That’s very close to Tesla’s charging rates with their Supercharger network.

What gas car people seem to forget is gas engines are efficient only in narrow RPM range and with full throttle. Diesel might not be as bad with regard to full open throttle that relate to pumping loss, but still maximum efficiency is not throughout all driving conditions. When they claim 40% (or whatever was latest trend), that’s only peak but typical would be far less, because the engine must operate in partial vacuum and sub-optimal RPM to reach 1 MPH to 100 MPH. Average might be who knows? 25%? That’s why hypermiling on gas car might involve pulse-and-glide.

EV also suffer the problem, but far less so than gas cars. In general, EV efficiency is power loss in higher current through resistive lossees, so slow going / slow regenerative braking would be most efficient beyond some fixed level (ie, power for CPU,, radio, etc), and pretty easy to hypermile.

“How could such a vehicle possibly be considered a supercar when it takes more than 12 seconds to reach 60 mph? Because once you’re there, it takes a scant 8.3 horsepower to maintain that speed — one-third that of a Jetta — and you can cruise along there all day while getting the equivalent of 261 mpg. That’s enough to go from San Francisco to Los Angeles and back on less than three gallons of fuel.”

SF to LA and back is about 760 miles, just to reinforce that Wired really does believe that the car can achieve 260 mpg in charge-sustaining mode.

I think that’s a load of crap, and that 120 mpg listed on wikipedia is a more representative charge-sustaining figure.

Perhaps an easier way to state this: a Jetta TDI which (according to Wired) requires 3x the power to drive at 60 mph returns far less than 1/3 of 260 mpg. One of these claims is not correct.

Regarding the rest: average thermal efficiency would depend greatly on the particular drive cycle being used. The ideal would be steady-state highway driving, at whatever load point hits the BSFC peak.

Resistive losses are relatively small in general for EVs, even if ignoring other platform losses and talking specifically about drivetrain efficiency. Cable runs are fairly short and they are sized for peak loads. When operated at typical highway speeds, current flow will be a fraction of peak (and since loss is I^2/R….)